Probability

1. Revision of probability

ID is: 3044 Seed is: 3935

Probability basics

Fill in the missing word or term for the sentence below:

Two events, $C$ and $D$ are $\underline{\quad \quad}$ if the outcomes of one event influences the outcome of the other.

NOTE: Remember that spelling is important! If you are unsure, look it up in your notes or textbook.

Answer:

Two events, $C$ and $D$ are if the outcomes of one event influences the outcome of the other.

one-of

type(string.nocase)

2 attempts remaining

STEP: <no title>
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For a list of terminology and definitions see the beginning of the Probability chapter in Everything Maths.

Two events, $C$ and $D$ are dependent if the outcomes of one event influences the outcome of the other.

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ID is: 3044 Seed is: 1730

Probability basics

Fill in the missing word or term for the sentence below:

The $\underline{\quad \quad}$ is the set of all possible outcomes of the experiment.

NOTE: Remember that spelling is important! If you are unsure, look it up in your notes or textbook.

Answer:

The is the set of all possible outcomes of the experiment.

one-of

type(string.nocase)

2 attempts remaining

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

For a list of terminology and definitions see the beginning of the Probability chapter in Everything Maths.

The sample space is the set of all possible outcomes of the experiment.

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ID is: 3044 Seed is: 4750

Probability basics

Fill in the missing word or term for the sentence below:

A probability of $\underline{\quad \quad}$ means the outcome of the experiment will never be in the event set.

NOTE: Remember that spelling is important! If you are unsure, look it up in your notes or textbook.

Answer:

A probability of means the outcome of the experiment will never be in the event set.

one-of

type(string.nocase)

2 attempts remaining

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

For a list of terminology and definitions see the beginning of the Probability chapter in Everything Maths.

A probability of 0 means the outcome of the experiment will never be in the event set.

Submit your answer as:

ID is: 3138 Seed is: 1396

Using the probability identity

Suppose we have two sets, $A$ and $B$ , which are in some sample space $S$ . Furthermore, suppose we know the following probabilities about events from these sets:

\begin{aligned} P (A) = \frac{1}{3} P (B) = \frac{1}{2} P (A and B) = \frac{1}{3} \end{aligned}

$\begin{aligned} P \left( A \right) = \frac{1}{3} \qquad P \left( B \right) = \frac{1}{2} \qquad P \left( A \text{ and } B \right) = \frac{1}{3} \end{aligned}$

Use the probability identity to calculate $P \left( A \text{ or } B \right)$ .

INSTRUCTION: Give an exact answer (do not round off).

Answer:

P (A or B) =

$P \left( A \text{ or } B \right) =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the probability identity. If you do not remember it, you can find it here.

STEP: Substitute the values into the probablity identity and work out the answer
[−2 points ⇒ 0 / 2 points left]

This question is about the probability identity:

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c}P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\end{array}$

This formula applies to any two sets inside of a sample space.

We have the three probabilities on the right hand side of the equation, so we just need to substitute in the values from the question and solve for $P \left( A \text{ or } B \right)$ .

\begin{aligned} P (A or B) & = (\frac{1}{3}) + (\frac{1}{2}) - (\frac{1}{3}) \\ = \frac{1}{2} \end{aligned}

$\begin{aligned} P \left( A \text{ or } B \right) & = \left( \frac{1}{3} \right) + \left( \frac{1}{2} \right) - \left( \frac{1}{3} \right) \\ & = \frac{1}{2} \end{aligned}$

The answer is $P \left( A \text{ or } B \right) = \dfrac{1}{2}$ .

Submit your answer as:

ID is: 3138 Seed is: 1864

Using the probability identity

Suppose we have two sets, $A$ and $B$ , which are in some sample space $S$ . Furthermore, suppose we know the following probabilities about events from these sets:

\begin{aligned} P (A) = \frac{1}{4} P (B) = \frac{7}{12} P (A and B) = \frac{1}{4} \end{aligned}

$\begin{aligned} P \left( A \right) = \frac{1}{4} \qquad P \left( B \right) = \frac{7}{12} \qquad P \left( A \text{ and } B \right) = \frac{1}{4} \end{aligned}$

Use the probability identity to calculate $P \left( A \text{ or } B \right)$ .

INSTRUCTION: Give an exact answer (do not round off).

Answer:

P (A or B) =

$P \left( A \text{ or } B \right) =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the probability identity. If you do not remember it, you can find it here.

STEP: Substitute the values into the probablity identity and work out the answer
[−2 points ⇒ 0 / 2 points left]

This question is about the probability identity:

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c}P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\end{array}$

This formula applies to any two sets inside of a sample space.

We have the three probabilities on the right hand side of the equation, so we just need to substitute in the values from the question and solve for $P \left( A \text{ or } B \right)$ .

\begin{aligned} P (A or B) & = (\frac{1}{4}) + (\frac{7}{12}) - (\frac{1}{4}) \\ = \frac{7}{12} \end{aligned}

$\begin{aligned} P \left( A \text{ or } B \right) & = \left( \frac{1}{4} \right) + \left( \frac{7}{12} \right) - \left( \frac{1}{4} \right) \\ & = \frac{7}{12} \end{aligned}$

The answer is $P \left( A \text{ or } B \right) = \dfrac{7}{12}$ .

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ID is: 3138 Seed is: 5520

Using the probability identity

Suppose we have two sets, $K$ and $L$ , which are in some sample space $S$ . Furthermore, suppose we know the following probabilities about events from these sets:

\begin{aligned} P (K) = \frac{3}{7} P (L) = \frac{3}{7} P (K and L) = \frac{2}{7} \end{aligned}

$\begin{aligned} P \left( K \right) = \frac{3}{7} \qquad P \left( L \right) = \frac{3}{7} \qquad P \left( K \text{ and } L \right) = \frac{2}{7} \end{aligned}$

Use the probability identity to calculate $P \left( K \text{ or } L \right)$ .

INSTRUCTION: Give an exact answer (do not round off).

Answer:

P (K or L) =

$P \left( K \text{ or } L \right) =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the probability identity. If you do not remember it, you can find it here.

STEP: Substitute the values into the probablity identity and work out the answer
[−2 points ⇒ 0 / 2 points left]

This question is about the probability identity:

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c}P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\end{array}$

This formula applies to any two sets inside of a sample space.

We have the three probabilities on the right hand side of the equation, so we just need to substitute in the values from the question and solve for $P \left( K \text{ or } L \right)$ .

\begin{aligned} P (K or L) & = (\frac{3}{7}) + (\frac{3}{7}) - (\frac{2}{7}) \\ = \frac{4}{7} \end{aligned}

$\begin{aligned} P \left( K \text{ or } L \right) & = \left( \frac{3}{7} \right) + \left( \frac{3}{7} \right) - \left( \frac{2}{7} \right) \\ & = \frac{4}{7} \end{aligned}$

The answer is $P \left( K \text{ or } L \right) = \dfrac{4}{7}$ .

Submit your answer as:

ID is: 1194 Seed is: 5105

Exploring definitions

A student wants to understand the term "experiment". So the student stands looking at a busy highway. Which of the following is the most appropriate example of the term "experiment"?

Answer: The most appropriate choice is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look up information about this in your notes or a book.

STEP: Select the best option from the choices in the list
[−1 point ⇒ 0 / 1 points left]

We recall the definition of the term "experiment":

$\Rightarrow$ An experiment refers to an uncertain process

Therefore the most appropriate example of the term "experiment" is will a truck drive past? . The other choices are also outcomes but not related to this specific experiment, therefore they are incorrect.

Submit your answer as:

ID is: 1194 Seed is: 2248

Exploring definitions

A student wants to understand the term "outcome". So the student rolls a die. Which of the following is the most appropriate example of the term "outcome"?

Answer: The most appropriate choice is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look up information about this in your notes or a book.

STEP: Select the best option from the choices in the list
[−1 point ⇒ 0 / 1 points left]

We recall the definition of the term "outcome":

$\Rightarrow$ An outcome of an experiment is a single result of that experiment

Therefore the most appropriate example of the term "outcome" is the die lands on the number 6 . The other choices are also outcomes but not related to this specific experiment, therefore they are incorrect.

Submit your answer as:

ID is: 1194 Seed is: 2034

Exploring definitions

A student wants to understand the term "experiment". So the student stands looking at a busy highway. Which of the following is the most appropriate example of the term "experiment"?

Answer: The most appropriate choice is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look up information about this in your notes or a book.

STEP: Select the best option from the choices in the list
[−1 point ⇒ 0 / 1 points left]

We recall the definition of the term "experiment":

$\Rightarrow$ An experiment refers to an uncertain process

Therefore the most appropriate example of the term "experiment" is will there be an accident? . The other choices are also outcomes but not related to this specific experiment, therefore they are incorrect.

Submit your answer as:

ID is: 3135 Seed is: 9325

Relative frequency compared to theoretical probability

Tshegofatso just flipped a coin $\text{3}$ times. The results of the flips were:

Flip 1	Flip 2	Flip 3	Next Flip
Tails	Heads	Heads	?

Answer the three questions which follow about this situation.

What is the most likely outcome of her next flip?

Answer: For the next flip, .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The results of the first $\text{3}$ flips do not influence the next flip.

STEP: Use the theoretical probability to determine the answer
[−1 point ⇒ 0 / 1 points left]

In this question, Tshegofatso is flipping a coin. She already flipped the coin $\text{3}$ times. From the list of results we can see that she got heads $\text{2}$ times and tails $\text{1}$ time. The question asks us what is likely to happen in the next flip.

The key is to realise that the past outcomes do not change the theoretical probability of the next flip. The theoretical probability for heads and tails is $\frac{1}{2}$ and that does not change based on earlier outcomes. Since there are more heads than tails in the first $\text{3}$ flips, it is tempting to think that the next flip is likely to be tails. But that is not true. Even if someone flips a coin and gets heads $\text{10}$ times in a row, the theoretical probability is still $\frac{1}{2}$ that the next flip will be heads again.

The correct choice from the list is: both outcomes are equally likely.

Submit your answer as:
What is the relative frequency of tails for the first $\text{3}$ flips? Give your answer as a fraction (not a percentage or a decimal).

Answer: The relative frequency of tails is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Relative frequency is the number of times a specific outcome happened divided by the total number of trials. You need to do that calculation using the information about the coin flips.

STEP: Calculate the relative frequency based on the results of the flips
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We need to calculate the relative frequency of tails for the first $\text{3}$ flips. Relative frequency is the ratio of how many times something happens to the total number of times we tried the experiment. In this case, that means the ratio of how many times Tshegofatso got tails to the total number of flips. This is:

$\begin{aligned} \text{Relative frequency of tails } \normalsize & = \frac{\text{number of tails}}{\text{number of flips}} \\ & = \frac{\text{1}}{\text{3}} \end{aligned}$

The relative frequency of tails from the first 3 flips is $\frac{1}{3}$ .

Submit your answer as:

The theoretical probability for getting tails for any flip is $\frac{1}{2}$ . However, the relative frequency in Question 2 was not equal to $\frac{1}{2}$ . How is this possible? Select your answer from the choices below.

A	The coin must be an unfair one.
B	Tshegofatso can not flip coins properly.
C	The theoretical probability must be wrong.
D	Relative frequency often does not equal theoretical probability.
E	None of the above

Answer: It is possible because .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Theoretical probability applies to each and every individual coin flip. The relative frequency summarises the outcome of an experiment. Must they always be the same?

STEP: Distinguish between relative frequency and theoretical probability
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This question is about the relationship between theoretical probability and relative frequency. The relative frequency depends on the sequence of outcomes that we observe while doing an experiment like flipping coins many times. The relative frequency can be different every time we redo the experiment. This is actually what probability means: the outcome is not predictable. When flipping a coin, it might land on heads and it might land on tails.

Theoretical probability tells us how likely the different outcomes are for a single event. For a coin flip, getting heads or tails are equally likely results. So the probability for heads and tails is $\frac{1}{2}$ each.

However, each flip is independent of the flips before and after it. As a result, it is common that the relative frequency of an event does not exactly equal the theoretical probability of the event. But the more times we flip the coin, the closer and closer the relative frequency gets to the theoretical probability.

The correct statement from the list is: Relative frequency often does not equal theoretical probability. So the correct choice in the table is D.

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ID is: 3135 Seed is: 1347

Relative frequency compared to theoretical probability

Lidija just flipped a coin $\text{3}$ times. The results of the flips were:

Flip 1	Flip 2	Flip 3	Next Flip
Tails	Tails	Heads	?

Answer the three questions which follow about this situation.

What is the most likely outcome of her next flip?

Answer: For the next flip, .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The results of the first $\text{3}$ flips do not influence the next flip.

STEP: Use the theoretical probability to determine the answer
[−1 point ⇒ 0 / 1 points left]

In this question, Lidija is flipping a coin. She already flipped the coin $\text{3}$ times. From the list of results we can see that she got heads $\text{1}$ time and tails $\text{2}$ times. The question asks us what is likely to happen in the next flip.

The key is to realise that the past outcomes do not change the theoretical probability of the next flip. The theoretical probability for heads and tails is $\frac{1}{2}$ and that does not change based on earlier outcomes. Since there are more tails than heads in the first $\text{3}$ flips, it is tempting to think that the next flip is likely to be heads. But that is not true. Even if someone flips a coin and gets tails $\text{10}$ times in a row, the theoretical probability is still $\frac{1}{2}$ that the next flip will be tails again.

The correct choice from the list is: both outcomes are equally likely.

Submit your answer as:
What is the relative frequency of heads for the first $\text{3}$ flips? Give your answer as a fraction (not a percentage or a decimal).

Answer: The relative frequency of heads is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Relative frequency is the number of times a specific outcome happened divided by the total number of trials. You need to do that calculation using the information about the coin flips.

STEP: Calculate the relative frequency based on the results of the flips
[−1 point ⇒ 0 / 1 points left]

We need to calculate the relative frequency of heads for the first $\text{3}$ flips. Relative frequency is the ratio of how many times something happens to the total number of times we tried the experiment. In this case, that means the ratio of how many times Lidija got heads to the total number of flips. This is:

$\begin{aligned} \text{Relative frequency of heads } \normalsize & = \frac{\text{number of heads}}{\text{number of flips}} \\ & = \frac{\text{1}}{\text{3}} \end{aligned}$

The relative frequency of heads from the first 3 flips is $\frac{1}{3}$ .

Submit your answer as:

The theoretical probability for getting heads for any flip is $\frac{1}{2}$ . However, the relative frequency in Question 2 was not equal to $\frac{1}{2}$ . What will happen to the relative frequency if Lidija keeps flipping the coin? Select your answer from the choices below.

A	Lidija might miss the bus home.
B	The pattern of more tails than heads will continue.
C	The relative frequency will get closer and closer to one half.
D	The relative frequency will get closer and closer to $\text{0}$ .
E	None of the above

Answer: If she keeps flipping, .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The more times Lidija flips the coin, the more you can expect the number of heads and tails to balance. If that happens, what will happen to the relative frequency?

STEP: Distinguish between relative frequency and theoretical probability
[−1 point ⇒ 0 / 1 points left]

The correct statement from the list is: The relative frequency will get closer and closer to one half. So the correct choice in the table is C.

Submit your answer as:

ID is: 3135 Seed is: 1305

Relative frequency compared to theoretical probability

Sindisiwe just flipped a coin $\text{4}$ times. The results of the flips were:

Flip 1	Flip 2	Flip 3	Flip 4	Next Flip
Tails	Heads	Heads	Heads	?

Answer the three questions which follow about this situation.

What is the most likely outcome of her next flip?

Answer: For the next flip, .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The results of the first $\text{4}$ flips do not influence the next flip.

STEP: Use the theoretical probability to determine the answer
[−1 point ⇒ 0 / 1 points left]

In this question, Sindisiwe is flipping a coin. She already flipped the coin $\text{4}$ times. From the list of results we can see that she got heads $\text{3}$ times and tails $\text{1}$ time. The question asks us what is likely to happen in the next flip.

The key is to realise that the past outcomes do not change the theoretical probability of the next flip. The theoretical probability for heads and tails is $\frac{1}{2}$ and that does not change based on earlier outcomes. Since there are more heads than tails in the first $\text{4}$ flips, it is tempting to think that the next flip is likely to be tails. But that is not true. Even if someone flips a coin and gets heads $\text{10}$ times in a row, the theoretical probability is still $\frac{1}{2}$ that the next flip will be heads again.

The correct choice from the list is: both outcomes are equally likely.

Submit your answer as:
What is the relative frequency of heads for the first $\text{4}$ flips? Give your answer as a fraction (not a percentage or a decimal).

Answer: The relative frequency of heads is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Relative frequency is the number of times a specific outcome happened divided by the total number of trials. You need to do that calculation using the information about the coin flips.

STEP: Calculate the relative frequency based on the results of the flips
[−1 point ⇒ 0 / 1 points left]

We need to calculate the relative frequency of heads for the first $\text{4}$ flips. Relative frequency is the ratio of how many times something happens to the total number of times we tried the experiment. In this case, that means the ratio of how many times Sindisiwe got heads to the total number of flips. This is:

$\begin{aligned} \text{Relative frequency of heads } \normalsize & = \frac{\text{number of heads}}{\text{number of flips}} \\ & = \frac{\text{3}}{\text{4}} \end{aligned}$

The relative frequency of heads from the first 4 flips is $\frac{3}{4}$ .

Submit your answer as:

The theoretical probability for getting heads for any flip is $\frac{1}{2}$ . However, the relative frequency in Question 2 was not equal to $\frac{1}{2}$ . What will happen to the relative frequency if Sindisiwe keeps flipping the coin? Select your answer from the choices below.

A	The relative frequency will get closer and closer to one half.
B	The pattern of more heads than tails will continue.
C	The relative frequency will get closer and closer to $\text{1}$ .
D	She will probably lose her coin.
E	None of the above

Answer: If she keeps flipping, .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The more times Sindisiwe flips the coin, the more you can expect the number of heads and tails to balance. If that happens, what will happen to the relative frequency?

STEP: Distinguish between relative frequency and theoretical probability
[−1 point ⇒ 0 / 1 points left]

The correct statement from the list is: The relative frequency will get closer and closer to one half. So the correct choice in the table is A.

Submit your answer as:

ID is: 1188 Seed is: 2041

Theoretical probability

A student finds a 6 sided die and then rolls the die once on a table. What is the probability that the die lands on either 4 or 5?

INSTRUCTION: Write your answer as a decimal (correct to 2 decimal places).

Answer: The probability is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must use the formula for probability:

\begin{matrix} P (E) = \frac{n (E)}{n (S)} \end{matrix}

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

STEP: Use the formula for theoretical probability
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

\begin{matrix} P (E) = \frac{n (E)}{n (S)} \end{matrix}

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

Secondly we identify the values for the situation described in the question:

\begin{aligned} n (E) & = number of outcomes in the event set = 2 \\ n (S) & = number of possible outcomes in the sample space = 6 \end{aligned}

$\begin{aligned} n(E) & = \text{number of outcomes in the event set} = \text{2} \\ n(S) & = \text{number of possible outcomes in the sample space} = \text{6} \end{aligned}$

Finally, we substitute into the formula and work out the answer.

\begin{aligned} P (E) & = \frac{n (E)}{n (S)} \\ = \frac{2}{6} \\ P (E) & \approx 0.3333 \end{aligned}

$\begin{aligned} P(E) & = \frac{n(E)}{n(S)} \\ & = \frac{\text{2}}{\text{6}} \\ P(E) & \approx \text{0.3333}\text{} \end{aligned}$

Remember that the question says your answer should be "correct to 2 decimal places," which means rounding to two places. Therefore, the probability that the die lands on either 4 or 5 $\approx \text{0.33} \text{}$ .

Submit your answer as:

ID is: 1188 Seed is: 3875

Theoretical probability

A student finds a deck of 52 cards and then takes one card from the deck. What is the probability that the card's suit is Spades?

INSTRUCTION: Write your answer as a simplified fraction.

Answer: The probability is .

fraction

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must use the formula for probability:

\begin{matrix} P (E) = \frac{n (E)}{n (S)} \end{matrix}

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

STEP: Use the formula for theoretical probability
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

\begin{matrix} P (E) = \frac{n (E)}{n (S)} \end{matrix}

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

Secondly we identify the values for the situation described in the question:

\begin{aligned} n (E) & = number of outcomes in the event set = 13 \\ n (S) & = number of possible outcomes in the sample space = 52 \end{aligned}

$\begin{aligned} n(E) & = \text{number of outcomes in the event set} = \text{13} \\ n(S) & = \text{number of possible outcomes in the sample space} = \text{52} \end{aligned}$

Finally, we substitute into the formula and work out the answer.

\begin{aligned} P (E) & = \frac{n (E)}{n (S)} \\ = \frac{13}{52} \\ P (E) & = \frac{1}{4} \end{aligned}

$\begin{aligned} P(E) & = \frac{n(E)}{n(S)} \\ & = \frac{\text{13}}{\text{52}} \\ P(E) & = \dfrac{1}{4}\text{} \end{aligned}$

Therefore, the probability that the card's suit is Spades $= \dfrac{1}{4} \text{}$ .

Submit your answer as:

ID is: 1188 Seed is: 5095

Theoretical probability

A student finds a 6 sided die and then rolls the die once on a table. What is the probability that the die lands on either 4 or 5?

INSTRUCTION: Write your answer as a percentage (rounded to the nearest whole number).

Answer: The probability is %.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must use the formula for probability:

\begin{matrix} P (E) = \frac{n (E)}{n (S)} \end{matrix}

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

STEP: Use the formula for theoretical probability
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

\begin{matrix} P (E) = \frac{n (E)}{n (S)} \end{matrix}

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

Secondly we identify the values for the situation described in the question:

\begin{aligned} n (E) & = number of outcomes in the event set = 2 \\ n (S) & = number of possible outcomes in the sample space = 6 \end{aligned}

$\begin{aligned} n(E) & = \text{number of outcomes in the event set} = \text{2} \\ n(S) & = \text{number of possible outcomes in the sample space} = \text{6} \end{aligned}$

Finally, we substitute into the formula and work out the answer.

\begin{aligned} P (E) & = \frac{n (E)}{n (S)} \\ = \frac{2}{6} \\ P (E) & \approx 33.33 % \end{aligned}

$\begin{aligned} P(E) & = \frac{n(E)}{n(S)} \\ & = \frac{\text{2}}{\text{6}} \\ P(E) & \approx 33.33\text{%} \end{aligned}$

Remember that the question says your answer should be "rounded to the nearest whole number." Therefore, the probability that the die lands on either 4 or 5 $\approx 33 \text{%}$ .

Submit your answer as:

ID is: 3137 Seed is: 2106

Applying theoretical probability to experimental trials

The theoretical probability for a special outcome is $\frac{1}{4}$ . Someone does an experiment about this event with $\text{16}$ trials.

Out of the $\text{16}$ trials, how many times should we expect to get the special outcome? In other words, for the $\text{16}$ trials, what is the most likely number of times we will get the special outcome?

Answer: We expect the special outcome to happen times.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The theoretical probability tells you how frequently you can expect to get the special outcome. So use the theoretical probability to get the answer.

STEP: Use the theoretical probability to get the answer
[−1 point ⇒ 0 / 1 points left]

This question tells us that the theoretical probability for some special outcome is $\frac{1}{4}$ . This means that we should expect the outcome to happen $\text{1}$ time out of every $\text{4}$ trials. For example, suppose the "special outcome" is picking a red marble from a sack of marbles. If we take a marble out of the sack $\text{4}$ times (putting the marble back each time), we will expect to get a red marble $\text{1}$ time.

For this question, there are $\text{16}$ trials. We expect $\frac{1}{4}$ of the trials to result in the special outcome:

$\begin{aligned} \text{special outcomes } \normalsize & = \text{(probability of the outcome)} \normalsize \cdot \text{(number of trials)} \\ \normalsize & = \frac{1}{4} \cdot \text{16} \\ & = \text{4} \end{aligned}$

Note: this calculation is related to the theoretical probability formula:

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

$P(E)$ is the theoretical probability. $n(S)$ is the number of trials (the total number of events in the sample space, no matter what the outcome is). $n(E)$ is the number of outcomes in the event set, which are the special outcomes:

$\begin{aligned} \frac{1}{4} & = \frac{\text{special outcomes }}{\text{number of trials}} \\ \normalsize \frac{1}{4} \cdot \text{(number of trials)} & = \text{special outcomes } \end{aligned}$

This is the same as the calculation we did above.

In the experiment with $\text{16}$ trials, we expect to get the special outcome $\text{4}$ times.

Submit your answer as:
In reality, the number of trials with the special outcome might not be the number from Question 1: it could be more or less. This is because the relative frequency does not always match the theoretical probability. For example, any of these four results is possible:

$\begin{aligned} 3& \text{ special outcomes from 16 trials} \normalsize\\10& \text{ special outcomes from 16 trials} \normalsize\\15& \text{ special outcomes from 16 trials} \normalsize\\16& \text{ special outcomes from 16 trials} \normalsize \end{aligned}$

Given that the theoretical probability is still $\frac{1}{4}$ , which of the four results listed above is the most likely?

Answer: The most likely result from the list is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The theoretical probability tells us that the most likely number of special outcomes is $\text{4}$ . The further you get from that number of special outcomes, the less likely it is.

STEP: Compare the choices in the list to the result of Question 1
[−1 point ⇒ 0 / 1 points left]

The question now asks us to identify the most likely number of special outcomes from a list of choices. We can answer this using the concept of theoretical probability. The theoretical probability tells us what the most likely number of special outcomes is. In Question 1 we found that this is $\text{4}$ of the $\text{16}$ trials. But of course the experiment might not result in $\text{4}$ special outcomes. That is because the outcomes are not predictable; they are probabilistic. For each individual trial, we cannot know if we will get the special outcome or not until we do the trial and see what we get.

So $\text{4}$ special outcomes is the most likely result of the experiment. But it is possible, and a little less likely, that we might get $\text{3}$ or $\text{5}$ special outcomes. It is also possible, though even less likely, that we might get $\text{2}$ or $\text{6}$ special outcomes. In fact, the further we get from $\text{4}$ special outcomes, the less likely that result becomes. This makes sense: if we expect $\text{4}$ special outcomes, it is quite unlikely that we will get the special outcome for all $\text{16}$ trials.

Now back to our list of choices. We need to identify the most likely number of special outcomes from a list of choices. The most likely option is the choice closest to the theoretically probable number of special outcomes.

So the most likely choice from the list is: 3 special outcomes for 16 trials.

Submit your answer as:

ID is: 3137 Seed is: 657

Applying theoretical probability to experimental trials

The theoretical probability for a special outcome is $\frac{1}{5}$ . Someone does an experiment about this event with $\text{15}$ trials.

Out of the $\text{15}$ trials, how many times should we expect to get the special outcome? In other words, for the $\text{15}$ trials, what is the most likely number of times we will get the special outcome?

Answer: We expect the special outcome to happen times.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The theoretical probability tells you how frequently you can expect to get the special outcome. So use the theoretical probability to get the answer.

STEP: Use the theoretical probability to get the answer
[−1 point ⇒ 0 / 1 points left]

This question tells us that the theoretical probability for some special outcome is $\frac{1}{5}$ . This means that we should expect the outcome to happen $\text{1}$ time out of every $\text{5}$ trials. For example, suppose the "special outcome" is picking a red marble from a sack of marbles. If we take a marble out of the sack $\text{5}$ times (putting the marble back each time), we will expect to get a red marble $\text{1}$ time.

For this question, there are $\text{15}$ trials. We expect $\frac{1}{5}$ of the trials to result in the special outcome:

$\begin{aligned} \text{special outcomes } \normalsize & = \text{(probability of the outcome)} \normalsize \cdot \text{(number of trials)} \\ \normalsize & = \frac{1}{5} \cdot \text{15} \\ & = \text{3} \end{aligned}$

Note: this calculation is related to the theoretical probability formula:

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

$P(E)$ is the theoretical probability. $n(S)$ is the number of trials (the total number of events in the sample space, no matter what the outcome is). $n(E)$ is the number of outcomes in the event set, which are the special outcomes:

$\begin{aligned} \frac{1}{5} & = \frac{\text{special outcomes }}{\text{number of trials}} \\ \normalsize \frac{1}{5} \cdot \text{(number of trials)} & = \text{special outcomes } \end{aligned}$

This is the same as the calculation we did above.

In the experiment with $\text{15}$ trials, we expect to get the special outcome $\text{3}$ times.

Submit your answer as:
In reality, the number of trials with the special outcome might not be the number from Question 1: it could be more or less. This is because the relative frequency does not always match the theoretical probability. For example, any of these four results is possible:

$\begin{aligned} 4& \text{ special outcomes from 15 trials} \normalsize\\11& \text{ special outcomes from 15 trials} \normalsize\\14& \text{ special outcomes from 15 trials} \normalsize\\15& \text{ special outcomes from 15 trials} \normalsize \end{aligned}$

Given that the theoretical probability is still $\frac{1}{5}$ , which of the four results listed above is the most likely?

Answer: The most likely result from the list is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The theoretical probability tells us that the most likely number of special outcomes is $\text{3}$ . The further you get from that number of special outcomes, the less likely it is.

STEP: Compare the choices in the list to the result of Question 1
[−1 point ⇒ 0 / 1 points left]

The question now asks us to identify the most likely number of special outcomes from a list of choices. We can answer this using the concept of theoretical probability. The theoretical probability tells us what the most likely number of special outcomes is. In Question 1 we found that this is $\text{3}$ of the $\text{15}$ trials. But of course the experiment might not result in $\text{3}$ special outcomes. That is because the outcomes are not predictable; they are probabilistic. For each individual trial, we cannot know if we will get the special outcome or not until we do the trial and see what we get.

So $\text{3}$ special outcomes is the most likely result of the experiment. But it is possible, and a little less likely, that we might get $\text{2}$ or $\text{4}$ special outcomes. It is also possible, though even less likely, that we might get $\text{1}$ or $\text{5}$ special outcomes. In fact, the further we get from $\text{3}$ special outcomes, the less likely that result becomes. This makes sense: if we expect $\text{3}$ special outcomes, it is quite unlikely that we will get the special outcome for all $\text{15}$ trials.

Now back to our list of choices. We need to identify the most likely number of special outcomes from a list of choices. The most likely option is the choice closest to the theoretically probable number of special outcomes.

So the most likely choice from the list is: 4 special outcomes for 15 trials.

Submit your answer as:

ID is: 3137 Seed is: 9749

Applying theoretical probability to experimental trials

The theoretical probability for a special outcome is $\frac{3}{4}$ . Someone does an experiment about this event with $\text{16}$ trials.

Out of the $\text{16}$ trials, how many times should we expect to get the special outcome? In other words, for the $\text{16}$ trials, what is the most likely number of times we will get the special outcome?

Answer: We expect the special outcome to happen times.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The theoretical probability tells you how frequently you can expect to get the special outcome. So use the theoretical probability to get the answer.

STEP: Use the theoretical probability to get the answer
[−1 point ⇒ 0 / 1 points left]

This question tells us that the theoretical probability for some special outcome is $\frac{3}{4}$ . This means that we should expect the outcome to happen $\text{3}$ times out of every $\text{4}$ trials. For example, suppose the "special outcome" is picking a red marble from a sack of marbles. If we take a marble out of the sack $\text{4}$ times (putting the marble back each time), we will expect to get a red marble $\text{3}$ times.

For this question, there are $\text{16}$ trials. We expect $\frac{3}{4}$ of the trials to result in the special outcome:

$\begin{aligned} \text{special outcomes } \normalsize & = \text{(probability of the outcome)} \normalsize \cdot \text{(number of trials)} \\ \normalsize & = \frac{3}{4} \cdot \text{16} \\ & = \text{12} \end{aligned}$

Note: this calculation is related to the theoretical probability formula:

$\begin{array}{c}P(E) = \frac{n(E)}{n(S)}\end{array}$

$P(E)$ is the theoretical probability. $n(S)$ is the number of trials (the total number of events in the sample space, no matter what the outcome is). $n(E)$ is the number of outcomes in the event set, which are the special outcomes:

$\begin{aligned} \frac{3}{4} & = \frac{\text{special outcomes }}{\text{number of trials}} \\ \normalsize \frac{3}{4} \cdot \text{(number of trials)} & = \text{special outcomes } \end{aligned}$

This is the same as the calculation we did above.

In the experiment with $\text{16}$ trials, we expect to get the special outcome $\text{12}$ times.

Submit your answer as:
In reality, the number of trials with the special outcome might not be the number from Question 1: it could be more or less. This is because the relative frequency does not always match the theoretical probability. For example, any of these four results is possible:

$\begin{aligned} 1 & \text{ special outcome from 16 trials} \normalsize\\3& \text{ special outcomes from 16 trials} \normalsize\\6& \text{ special outcomes from 16 trials} \normalsize\\13& \text{ special outcomes from 16 trials} \normalsize \end{aligned}$

Given that the theoretical probability is still $\frac{3}{4}$ , which of the four results listed above is the most likely?

Answer: The most likely result from the list is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The theoretical probability tells us that the most likely number of special outcomes is $\text{12}$ . The further you get from that number of special outcomes, the less likely it is.

STEP: Compare the choices in the list to the result of Question 1
[−1 point ⇒ 0 / 1 points left]

The question now asks us to identify the most likely number of special outcomes from a list of choices. We can answer this using the concept of theoretical probability. The theoretical probability tells us what the most likely number of special outcomes is. In Question 1 we found that this is $\text{12}$ of the $\text{16}$ trials. But of course the experiment might not result in $\text{12}$ special outcomes. That is because the outcomes are not predictable; they are probabilistic. For each individual trial, we cannot know if we will get the special outcome or not until we do the trial and see what we get.

So $\text{12}$ special outcomes is the most likely result of the experiment. But it is possible, and a little less likely, that we might get $\text{11}$ or $\text{13}$ special outcomes. It is also possible, though even less likely, that we might get $\text{10}$ or $\text{14}$ special outcomes. In fact, the further we get from $\text{12}$ special outcomes, the less likely that result becomes. This makes sense: if we expect $\text{12}$ special outcomes, it is quite unlikely that we will never get the special outcome at all (for none of the trials).

Now back to our list of choices. We need to identify the most likely number of special outcomes from a list of choices. The most likely option is the choice closest to the theoretically probable number of special outcomes.

So the most likely choice from the list is: 13 special outcomes for 16 trials.

Submit your answer as:

ID is: 1638 Seed is: 4104

Revision: calculations with probability

In a group of $\text{277}$ students $\text{130}$ take Natural Science and $\text{153}$ take Mathematics while $\text{38}$ chose neither subject.

If a student is selected at random, what is the probability that the student does not take Natural Science or Mathematics?

Give your answer as a number

INSTRUCTION: Round your answer to one decimal place.

Answer: The probability is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Construct a Venn Diagram which represents this information

STEP: Shade the relevant areas on the Venn diagram
[−2 points ⇒ 0 / 2 points left]

You can find the answer from the given information, how many students took neither subject?

The Venn diagram represents the event.

NOTE: Remember: ("and" is the same as intersection

\cap

$\cap$ ) and ("or" is the same as union

\cup

$\cup$ )

\begin{aligned} n (N or M) = n (N) + n (M) - (N and M) \end{aligned}

$\begin{aligned} n(N \text{ or } M) = n(N) + n(M) - (N \text{ and } M) \end{aligned}$

\begin{matrix} P (not N or M) = (\frac{38}{277}) \end{matrix}

$\begin{array}{c}P \left(\text{ not }N \text{ or } M \right) = \left(\frac{ 38}{ 277} \right)\end{array}$

The final answer is: $P = \frac{38}{277} = \text{0.1}$ .

Submit your answer as:

ID is: 1638 Seed is: 6403

Revision: calculations with probability

In a group of $\text{277}$ students $\text{149}$ take Natural Science and $\text{135}$ take Afrikaans while $\text{27}$ chose neither subject.

If a student is selected at random, what is the probability that the student takes Afrikaans only?

INSTRUCTION: Write your answer as a simplified fraction.

Answer: The probability is .

fraction

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Construct a Venn Diagram which represents this information

STEP: Shade the relevant areas on the Venn diagram
[−2 points ⇒ 0 / 2 points left]

This question refers to the Probabiltiy of Afrikaans written as $P$ (Afrikaans not Natural Science).

The Venn diagram represents the event.

NOTE: Remember: ("and" is the same as intersection

\cap

$\cap$ ) and ("or" is the same as union

\cup

$\cup$ )

The Venn diagram is divided into $\text{4}$ sections. The shaded area represents $P$ (Afrikaans not Natural Science). First complete all the given information on the diagram, then solve the missing parts. We follow these steps:

Step 1:
Union = total students - students that took neither subject
Union = (277) - (27) = 250

Step 2: Union = (Natural Science) + (Afrikaans) - intersection
Rearrange the formula:
Intersection = (Natural Science) + (Afrikaans) - union
Intersection = 34

Step 3: All of Afrikaans - intersection = Area of (Afrikaans not Natural Science)
(Afrikaans not Natural Science) = 101

\begin{matrix} P (A not N) = (\frac{101}{277}) \end{matrix}

$\begin{array}{c}P\left(A \text{ not } N\right) = \left(\frac{ 101}{ 277}\right)\end{array}$

The final answer is: $P = \frac{101}{277}$ .

Submit your answer as:

ID is: 1638 Seed is: 9166

Revision: calculations with probability

In a group of $\text{266}$ students $\text{111}$ take Natural Science and $\text{168}$ take Mathematics while $\text{36}$ chose neither subject.

If a student is selected at random, what is the probability that the student takes both Natural Science and Mathematics?

Give your answer as a number

INSTRUCTION: Round your answer to one decimal place.

Answer: The probability is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Construct a Venn Diagram which represents this information

STEP: Shade the relevant areas on the Venn diagram
[−3 points ⇒ 0 / 3 points left]

The question refers to the INTERSECTON of Natural Science and Mathematics.

The Venn diagram represents the event.

NOTE: Remember: ("and" is the same as intersection

\cap

$\cap$ ) and ("or" is the same as union

\cup

$\cup$ )

The Venn diagram is divided into 4 sections. Fill in the given information and calculate the missing section which is represented by the shaded area. In this case you want the amount of students taking N $\cap$ M. We follow these steps:

Step 1:
Union = total students - students that took neither subject
Union = (266) - (36) = 230

Step 2: Union = (Natural Science) + (Mathematics) - intersection
Rearrange the formula:
Intersection = (Natural Science) + (Mathematics) - union
Intersection = 49

\begin{aligned} n (N or M) = n (N) + n (M) - n (N and M) \end{aligned}

$\begin{aligned} n\left(N \text{ or } M\right)= n\left(N\right)+n\left(M\right) - n\left(N \text{ and } M\right) \end{aligned}$

\begin{matrix} ∴ n (N \cup M) = n (N) + n (M) - n (N \cap M) \end{matrix}

$\begin{array}{c}\therefore n\left(N \cup M\right)= n\left(N\right)+n\left(M\right) - n\left(N \cap M\right)\end{array}$

\begin{matrix} n (N \cap M) = n (N) + n (M) - n (N \cup M) \end{matrix}

$\begin{array}{c}n\left(N \cap M\right)= n\left(N\right)+n\left(M\right) - n\left(N \cup M\right)\end{array}$

\begin{matrix} (N \cap M) = 111 + 168 - 230 \end{matrix}

$\begin{array}{c}(N \cap M) = 111 + 168 - 230\end{array}$

\begin{matrix} P (N \cap M) = \frac{49}{266} \end{matrix}

$\begin{array}{c}P\left(N∩ M\right)= \frac{ 49}{ 266}\end{array}$

The final answer is: $P = \frac{7}{38} = \text{0.2}$ .

Submit your answer as:

ID is: 1189 Seed is: 8013

Relative frequency

A die is tossed 35 times and lands 4 times on the number 5.

What is the relative frequency of observing the die land on the number 5? Write your answer correct to 2 decimal places.

Answer: The relative frequency is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by writing down the formula for the relative frequency, $f = \frac{p}{t}$ . Then figure out where to substitute the numbers in the question.

STEP: Use the values given with the formula for the relative frequency
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

\begin{matrix} f = \frac{p}{t} \end{matrix}

$\begin{array}{c}f = \frac{p}{t}\end{array}$

Secondly we identify variables needed:

\begin{aligned} p & = number of positive trials = 4 \\ t & = total number of trials = 35 \end{aligned}

$\begin{aligned} p & = \text{number of positive trials} = \text{4} \\ t & = \text{total number of trials} = \text{35} \end{aligned}$

Finally, we solve the equation:

\begin{aligned} f & = \frac{p}{t} \\ = \frac{4}{35} \\ = 0.11 \end{aligned}

$\begin{aligned} f & = \frac{p}{t} \\ & = \frac{\text{4}}{\text{35}} \\ & = \text{0.11} \end{aligned}$

Therefore, the relative frequency of observing the die on the number 5 is $\text{0.11}$ .

Submit your answer as:

ID is: 1189 Seed is: 1055

Relative frequency

A die is tossed 26 times and lands 4 times on the number 3.

What is the relative frequency of observing the die land on the number 3? Write your answer correct to 2 decimal places.

Answer: The relative frequency is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by writing down the formula for the relative frequency, $f = \frac{p}{t}$ . Then figure out where to substitute the numbers in the question.

STEP: Use the values given with the formula for the relative frequency
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

\begin{matrix} f = \frac{p}{t} \end{matrix}

$\begin{array}{c}f = \frac{p}{t}\end{array}$

Secondly we identify variables needed:

\begin{aligned} p & = number of positive trials = 4 \\ t & = total number of trials = 26 \end{aligned}

$\begin{aligned} p & = \text{number of positive trials} = \text{4} \\ t & = \text{total number of trials} = \text{26} \end{aligned}$

Finally, we solve the equation:

\begin{aligned} f & = \frac{p}{t} \\ = \frac{4}{26} \\ = 0.15 \end{aligned}

$\begin{aligned} f & = \frac{p}{t} \\ & = \frac{\text{4}}{\text{26}} \\ & = \text{0.15} \end{aligned}$

Therefore, the relative frequency of observing the die on the number 3 is $\text{0.15}$ .

Submit your answer as:

ID is: 1189 Seed is: 9927

Relative frequency

A coin is tossed 41 times and lands 20 times on tails.

What is the relative frequency of observing the coin land on tails? Write your answer correct to 2 decimal places.

Answer: The relative frequency is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by writing down the formula for the relative frequency, $f = \frac{p}{t}$ . Then figure out where to substitute the numbers in the question.

STEP: Use the values given with the formula for the relative frequency
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

\begin{matrix} f = \frac{p}{t} \end{matrix}

$\begin{array}{c}f = \frac{p}{t}\end{array}$

Secondly we identify variables needed:

\begin{aligned} p & = number of positive trials = 20 \\ t & = total number of trials = 41 \end{aligned}

$\begin{aligned} p & = \text{number of positive trials} = \text{20} \\ t & = \text{total number of trials} = \text{41} \end{aligned}$

Finally, we solve the equation:

\begin{aligned} f & = \frac{p}{t} \\ = \frac{20}{41} \\ = 0.49 \end{aligned}

$\begin{aligned} f & = \frac{p}{t} \\ & = \frac{\text{20}}{\text{41}} \\ & = \text{0.49} \end{aligned}$

Therefore, the relative frequency of observing the coin on tails is $\text{0.49}$ .

Submit your answer as:

ID is: 1187 Seed is: 19

Probability identities

A group of students is given the following event sets:

Event Set

B

$B$

1

$\text{1}$

2

$\text{2}$

3

$\text{3}$

5

$\text{5}$

6

$\text{6}$

Event Set

(A or B)

$\left(A \text{ or } B\right)$

1

$\text{1}$

2

$\text{2}$

3

$\text{3}$

5

$\text{5}$

6

$\text{6}$

Event Set

(A and B)

$\left(A \text{ and } B\right)$

2

$\text{2}$

3

$\text{3}$

5

$\text{5}$

6

$\text{6}$

The sample space is $\{ 1 ;\; 2 ;\; 3 ;\; 4 ;\; 5 ;\; 6 \}$ .

They are asked to calculate the value of $P(A)$ . They get stuck, and you offer to calculate it for them.

Answer:

P (A)

$P(A)$ =

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down the probability identity. If you are not sure what that is, you can find a review here.

STEP: Rearrange the probabilty identity for the value we want
[−1 point ⇒ 2 / 3 points left]

To calculate probability, we use the probability identity:

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c}P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\end{array}$

Rearranging the formula to make $P(A)$ the subject, we get:

\begin{aligned} P (A) & = P (A or B) - P (B) + P (A and B) \end{aligned}

$\begin{aligned} P(A) & = P(A \text{ or } B) - P(B) + P(A \text{ and } B) \end{aligned}$

STEP: Calculate each probability that we need
[−1 point ⇒ 1 / 3 points left]

Now we need to identify variables needed:

\begin{aligned} P (B) & = \frac{n (B)}{n (S)} = \frac{5}{6} \\ P (A or B) & = \frac{n (A or B)}{n (S)} = \frac{5}{6} \\ P (A and B) & = \frac{n (A and B)}{n (S)} = \frac{4}{6} \end{aligned}

$\begin{aligned} P(B) & = \frac{n(B)}{n(S)} = \frac{5}{6} \\ P(A \text{ or } B) & = \frac{n(A \text{ or } B)}{n(S)} = \frac{5}{6} \\ P(A \text{ and } B) & = \frac{n(A \text{ and } B)}{n(S)} = \frac{4}{6} \end{aligned}$

STEP: Substitute in the values we know and evaluate
[−1 point ⇒ 0 / 3 points left]

Finally, substitute in and solve for the unknown value $P(A)$ :

\begin{aligned} P (A) & = P (A or B) - P (B) + P (A and B) \\ P (A) & = (\frac{5}{6}) - (\frac{5}{6}) + (\frac{4}{6}) \\ P (A) & = \frac{2}{3} \end{aligned}

$\begin{aligned} P(A) & = P(A \text{ or } B) - P(B) + P(A \text{ and } B) \\ P(A) & = \left( \frac{5}{6} \right) - \left( \frac{5}{6} \right) + \left( \frac{4}{6} \right) \\ P(A) & = \frac{2}{3} \end{aligned}$

Therefore, the value of $P(A)$ is $\frac{2}{3}$ .

Submit your answer as:

ID is: 1187 Seed is: 1499

Probability identities

A group of students is given the following event sets:

Event Set

B

$B$

1

$\text{1}$

2

$\text{2}$

3

$\text{3}$

4

$\text{4}$

6

$\text{6}$

Event Set

(A or B)

$\left(A \text{ or } B\right)$

1

$\text{1}$

2

$\text{2}$

3

$\text{3}$

4

$\text{4}$

6

$\text{6}$

Event Set

(A and B)

$\left(A \text{ and } B\right)$

2

$\text{2}$

3

$\text{3}$

The sample space is $\{ 1 ;\; 2 ;\; 3 ;\; 4 ;\; 5 ;\; 6 \}$ .

They are asked to calculate the value of $P(A)$ . They get stuck, and you offer to calculate it for them.

Answer:

P (A)

$P(A)$ =

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down the probability identity. If you are not sure what that is, you can find a review here.

STEP: Rearrange the probabilty identity for the value we want
[−1 point ⇒ 2 / 3 points left]

To calculate probability, we use the probability identity:

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c}P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\end{array}$

Rearranging the formula to make $P(A)$ the subject, we get:

\begin{aligned} P (A) & = P (A or B) - P (B) + P (A and B) \end{aligned}

$\begin{aligned} P(A) & = P(A \text{ or } B) - P(B) + P(A \text{ and } B) \end{aligned}$

STEP: Calculate each probability that we need
[−1 point ⇒ 1 / 3 points left]

Now we need to identify variables needed:

\begin{aligned} P (B) & = \frac{n (B)}{n (S)} = \frac{5}{6} \\ P (A or B) & = \frac{n (A or B)}{n (S)} = \frac{5}{6} \\ P (A and B) & = \frac{n (A and B)}{n (S)} = \frac{2}{6} \end{aligned}

STEP: Substitute in the values we know and evaluate
[−1 point ⇒ 0 / 3 points left]

Finally, substitute in and solve for the unknown value $P(A)$ :

\begin{aligned} P (A) & = P (A or B) - P (B) + P (A and B) \\ P (A) & = (\frac{5}{6}) - (\frac{5}{6}) + (\frac{2}{6}) \\ P (A) & = \frac{1}{3} \end{aligned}

$\begin{aligned} P(A) & = P(A \text{ or } B) - P(B) + P(A \text{ and } B) \\ P(A) & = \left( \frac{5}{6} \right) - \left( \frac{5}{6} \right) + \left( \frac{2}{6} \right) \\ P(A) & = \frac{1}{3} \end{aligned}$

Therefore, the value of $P(A)$ is $\frac{1}{3}$ .

Submit your answer as:

ID is: 1187 Seed is: 6509

Probability identities

A group of students is given the following event sets:

Event Set

B

$B$

5

$\text{5}$

Event Set

(A or B)

$\left(A \text{ or } B\right)$

1

$\text{1}$

5

$\text{5}$

6

$\text{6}$

Event Set

(A and B)

$\left(A \text{ and } B\right)$

empty

The sample space is $\{ 1 ;\; 2 ;\; 3 ;\; 4 ;\; 5 ;\; 6 \}$ .

They are asked to calculate the value of $P(A)$ . They get stuck, and you offer to calculate it for them.

Answer:

P (A)

$P(A)$ =

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down the probability identity. If you are not sure what that is, you can find a review here.

STEP: Rearrange the probabilty identity for the value we want
[−1 point ⇒ 2 / 3 points left]

To calculate probability, we use the probability identity:

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c}P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\end{array}$

Rearranging the formula to make $P(A)$ the subject, we get:

\begin{aligned} P (A) & = P (A or B) - P (B) + P (A and B) \end{aligned}

$\begin{aligned} P(A) & = P(A \text{ or } B) - P(B) + P(A \text{ and } B) \end{aligned}$

STEP: Calculate each probability that we need
[−1 point ⇒ 1 / 3 points left]

Now we need to identify variables needed:

\begin{aligned} P (B) & = \frac{n (B)}{n (S)} = \frac{1}{6} \\ P (A or B) & = \frac{n (A or B)}{n (S)} = \frac{3}{6} \\ P (A and B) & = \frac{n (A and B)}{n (S)} = \frac{0}{6} \end{aligned}

$\begin{aligned} P(B) & = \frac{n(B)}{n(S)} = \frac{1}{6} \\ P(A \text{ or } B) & = \frac{n(A \text{ or } B)}{n(S)} = \frac{3}{6} \\ P(A \text{ and } B) & = \frac{n(A \text{ and } B)}{n(S)} = \frac{0}{6} \end{aligned}$

STEP: Substitute in the values we know and evaluate
[−1 point ⇒ 0 / 3 points left]

Finally, substitute in and solve for the unknown value $P(A)$ :

\begin{aligned} P (A) & = P (A or B) - P (B) + P (A and B) \\ P (A) & = (\frac{3}{6}) - (\frac{1}{6}) + (\frac{0}{6}) \\ P (A) & = \frac{1}{3} \end{aligned}

$\begin{aligned} P(A) & = P(A \text{ or } B) - P(B) + P(A \text{ and } B) \\ P(A) & = \left( \frac{3}{6} \right) - \left( \frac{1}{6} \right) + \left( \frac{0}{6} \right) \\ P(A) & = \frac{1}{3} \end{aligned}$

Therefore, the value of $P(A)$ is $\frac{1}{3}$ .

Submit your answer as:

2. Equiprobable sample space

3. Mutually exclusive events

ID is: 1191 Seed is: 4075

Mutually exclusive events

Consider the following Venn diagram, which shows two sets in the sample space $\{ n:n \text{ } \epsilon \text{ } \mathbb{Z}, \text{ } 1 \leq n \leq 15 \}$ (the first $\text{15}$ natural numbers).

Are the following two sets mutually exclusive, or not?

\begin{array}{ccc} Set 1: & Set 2: \\ (A or B) & (A or B)^{'} \end{array}

$\begin{array}{ccc} \text{Set 1:} & & \text{Set 2:} \\ \hline (A \text{ or } B) & \quad & (A \text{ or } B)' \end{array}$

Answer: Are

(A or B)

$(A \text{ or } B)$ and

(A or B)^{'}

$(A \text{ or } B)'$ mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the two sets $(A \text{ or } B)$ and $(A \text{ or } B)'$ on the Venn diagram.

STEP: Identify the elements in the two sets and compare them
[−1 point ⇒ 0 / 1 points left]

We need to determine whether or not two sets are mutually exclusive. Two sets are called mutually exclusive if they do not share any elements (outcomes). To find the answer, we need to compare the elements in the two sets.

The event set for $(A \text{ or } B)$ is: $\{1;2;3;4;6;7;8;11;12;14;15\}$ .
The event set for $(A \text{ or } B)'$ is: $\{5;9;10;13\}$ .

The question we must ask: do the sets have any elements in common? By comparing the sets, we can identify the following overlapping event set: $\{\text{empty}\}$ .

Therefore the event sets $(A \text{ or } B)$ and $(A \text{ or } B)'$ are mutually exclusive in this example. The correct answer is Yes.

Submit your answer as:

ID is: 1191 Seed is: 1085

Mutually exclusive events

Consider the following Venn diagram, which shows two sets in the sample space $\{ n:n \text{ } \epsilon \text{ } \mathbb{Z}, \text{ } 1 \leq n \leq 15 \}$ (the first $\text{15}$ natural numbers).

Are the following two sets mutually exclusive, or not?

\begin{array}{ccc} Set 1: & Set 2: \\ (A and B) & (A or B)^{'} \end{array}

$\begin{array}{ccc} \text{Set 1:} & & \text{Set 2:} \\ \hline (A \text{ and } B) & \quad & (A \text{ or } B)' \end{array}$

Answer: Are

(A and B)

$(A \text{ and } B)$ and

(A or B)^{'}

$(A \text{ or } B)'$ mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the two sets $(A \text{ and } B)$ and $(A \text{ or } B)'$ on the Venn diagram.

STEP: Identify the elements in the two sets and compare them
[−1 point ⇒ 0 / 1 points left]

The event set for $(A \text{ and } B)$ is: $\{2;7;9;13;15\}$ .
The event set for $(A \text{ or } B)'$ is: $\{5;8;14\}$ .

The question we must ask: do the sets have any elements in common? By comparing the sets, we can identify the following overlapping event set: $\{\text{empty}\}$ .

Therefore the event sets $(A \text{ and } B)$ and $(A \text{ or } B)'$ are mutually exclusive in this example. The correct answer is Yes.

Submit your answer as:

ID is: 1191 Seed is: 482

Mutually exclusive events

Consider the following Venn diagram, which shows two sets in the sample space $\{ n:n \text{ } \epsilon \text{ } \mathbb{Z}, \text{ } 1 \leq n \leq 15 \}$ (the first $\text{15}$ natural numbers).

Are the following two sets mutually exclusive, or not?

\begin{array}{ccc} Set 1: & Set 2: \\ (A or B)^{'} & (A or B) \end{array}

$\begin{array}{ccc} \text{Set 1:} & & \text{Set 2:} \\ \hline (A \text{ or } B)' & \quad & (A \text{ or } B) \end{array}$

Answer: Are

(A or B)^{'}

$(A \text{ or } B)'$ and

(A or B)

$(A \text{ or } B)$ mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the two sets $(A \text{ or } B)'$ and $(A \text{ or } B)$ on the Venn diagram.

STEP: Identify the elements in the two sets and compare them
[−1 point ⇒ 0 / 1 points left]

The event set for $(A \text{ or } B)'$ is: $\{\}$ .
The event set for $(A \text{ or } B)$ is: $\{1;2;3;4;5;6;7;8;9;10;11;12;13;14;15\}$ .

The question we must ask: do the sets have any elements in common? By comparing the sets, we can identify the following overlapping event set: $\{\text{empty}\}$ .

Therefore the event sets $(A \text{ or } B)'$ and $(A \text{ or } B)$ are mutually exclusive in this example. The correct answer is Yes.

Submit your answer as:

ID is: 3166 Seed is: 7591

Mutually exclusive sets

You are given the following two sets:

\begin{aligned} A & = {46; 107; 98; 105} \\ B & = {105; 109; 112; 53} \end{aligned}

$\begin{aligned} \text{A} &= \left\{46 ; 107 ; 98 ; 105\right\} \\ \text{B} &= \left\{105 ; 109 ; 112 ; 53\right\} \end{aligned}$

From the drop down box below choose whether the values in these sets are mutually exclusive or not.

Answer: These sets are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive sets can never have common values.

STEP: Compare the values in the two given sets
[−1 point ⇒ 0 / 1 points left]

When sets are not mutually exclusive that means at least one element belongs to both sets at the same time. In this specific case an element that belongs in both sets is: $\text{105}$ .

The correct choice from the list is: not mutually exclusive.

Submit your answer as:

ID is: 3166 Seed is: 8862

Mutually exclusive sets

You are given the following two sets:

\begin{aligned} A & = {79; 97; 109; 93} \\ B & = {103; 49; 84; 96} \end{aligned}

$\begin{aligned} \text{A} &= \left\{79 ; 97 ; 109 ; 93\right\} \\ \text{B} &= \left\{103 ; 49 ; 84 ; 96\right\} \end{aligned}$

From the drop down box below choose whether the values in these sets are mutually exclusive or not.

Answer: These sets are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive sets can never have common values.

STEP: Compare the values in the two given sets
[−1 point ⇒ 0 / 1 points left]

Mutually exclusive sets mean that we cannot pick one element which belongs to both of the sets. So whatever number we pick, it cannot belong to both sets. If we compare the elements in these two sets, we find that not a single value belongs in both sets. This is what the diagram represents below.

The correct choice from the list is: mutually exclusive.

Submit your answer as:

ID is: 3166 Seed is: 2773

Mutually exclusive sets

You are given the following two sets:

\begin{aligned} A & = {101; 94; 103; 112} \\ B & = {113; 54; 58; 97} \end{aligned}

$\begin{aligned} \text{A} &= \left\{101 ; 94 ; 103 ; 112\right\} \\ \text{B} &= \left\{113 ; 54 ; 58 ; 97\right\} \end{aligned}$

From the drop down box below choose whether the values in these sets are mutually exclusive or not.

Answer: These sets are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive sets can never have common values.

STEP: Compare the values in the two given sets
[−1 point ⇒ 0 / 1 points left]

The correct choice from the list is: mutually exclusive.

Submit your answer as:

ID is: 4344 Seed is: 5403

Prove that two events are mutually exclusive

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

Given two events $A$ and $B$ :

$P(A) = \text{0.4}$
$P(B) = \text{0.5}$
$P(A \text{ or } B) = \text{0.8}$

Are the events $A$ and $B$ mutually exclusive? Justify your answer with appropriate calculations.

INSTRUCTION: Choose the most correct option to answer this question.

Answer:

Option 1
$\begin{aligned} P(A) + P(B) & = \text{0.9} \\ P(A \text{ and } B) & = \text{0.2} \\ \therefore P(A \text{ and } B) & \neq P(A) + P(B) \end{aligned}$

Therefore the events are not mutually exclusive.
Option 2
$\begin{aligned} P(A \text{ and } B) & = \text{0.2} \\ \therefore P(A \text{ and } B) & = P(A) \times P(B) \end{aligned}$

Therefore the events are mutually exclusive.
Option 3
$\begin{aligned} P(A) \times P(B) & = \text{0.2} \\ \therefore P(A) \times P(B) & \neq 0 \end{aligned}$

Therefore the events are not mutually exclusive.
Option 4
$\begin{aligned} P(A) + P(B) & = \text{0.9} \\ \therefore P(A) + P(B) & \neq P(A \text{ or } B) \end{aligned}$

Therefore the events are not mutually exclusive.

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

If $P(A \text{ and } B) = 0$ then two events are mutually exclusive.

STEP: How do we decide if events are mutually exclusive?
[−1 point ⇒ 3 / 4 points left]

Mutually exclusive events can never occur at the same time. For example, "Nneka is in the Western Cape" and "Nneka is in Gauteng" are mutually exclusive events, because Nneka cannot be in two places at once.

NOTE: Two events are mutually exclusive if the probability of both occuring at the same time is zero:

P (A and B) = 0

$P(A \text{ and } B) = 0$ .

STEP: Decide if the events are mutually exclusive
[−3 points ⇒ 0 / 4 points left]

For all events $A$ and $B$ :

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c} P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \end{array}$

There are two ways to prove that events are mutually exclusive:

Prove that $P(A \text{ and } B) = 0$ .
Prove that $P(A \text{ or } B) = P(A) + P(B)$ .

We know that:

${\color{red}{P(A) = \text{0.4}}}$
${\color{blue}{P(B) = \text{0.5}}}$ , and
${\color{ForestGreen}{P(A \text{ or } B) = \text{0.8}}}$ .

Method 1: Events $A$ and $B$ are mutually exclusive if $P(A \text{ and } B) = 0$

\begin{matrix} 0.8 = 0.4 + 0.5 - P (A and B) \end{matrix}

$\begin{array}{c} {\color{ForestGreen}{\text{0.8}}} = {\color{red}{\text{0.4}}} + {\color{blue}{\text{0.5}}} - P(A \text { and } B) \end{array}$

\begin{aligned} ∴ P (A and B) & = 0.1 \\ \neq 0 \end{aligned}

$\begin{aligned} \therefore P(A \text { and } B) & = \text{0.1} \\ & \neq 0 \end{aligned}$

Therefore the events are not mutually exclusive.

Method 2: Events $A$ and $B$ are mutually exclusive if $P(A \text{ or } B) = P(A) + P(B)$

\begin{aligned} P (A) + P (B) & = 0.4 + 0.5 \\ = 0.9 \\ ∴ P (A) + P (B) & \neq P (A or B) \end{aligned}

$\begin{aligned} P(A) + P(B) & = {\color{red}{\text{0.4}}} + {\color{blue}{\text{0.5}}} \\ & = \text{0.9} \\ \therefore P(A) + P(B) & \neq P(A \text{ or } B) \end{aligned}$

Therefore the events are not mutually exclusive.

NOTE: Choose the method that you prefer, and use that one. You don't need to use both methods!

Of the given choices, Option 4 is correct.

Submit your answer as:

Option 4

\begin{aligned} P (A) + P (B) & = 0.9 \\ ∴ P (A) + P (B) & \neq P (A or B) \end{aligned}

$\begin{aligned} P(A) + P(B) & = \text{0.9} \\ \therefore P(A) + P(B) & \neq P(A \text{ or } B) \end{aligned}$

Therefore the events are not mutually exclusive.

ID is: 4344 Seed is: 7696

Prove that two events are mutually exclusive

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

Given two events $A$ and $B$ :

$P(A) = \text{0.4}$
$P(B) = \text{0.3}$
$P(A \text{ or } B) = \text{0.7}$

Are the events $A$ and $B$ mutually exclusive? Justify your answer with appropriate calculations.

INSTRUCTION: Choose the most correct option to answer this question.

Answer:

Option 1
$\begin{aligned} P(A \text{ and } B) & = \text{0.12} \\ \therefore P(A \text{ and } B) & = P(A) \times P(B) \end{aligned}$

Therefore the events are mutually exclusive.
Option 2
$\begin{aligned} P(A) + P(B) & = \text{0.7} \\ \therefore P(A) + P(B) & = P(A \text{ or } B) \end{aligned}$

Therefore the events are mutually exclusive.
Option 3
$\begin{aligned} P(A) \times P(B) & = \text{0.12} \\ \therefore P(A) \times P(B) & \neq 0 \end{aligned}$

Therefore the events are not mutually exclusive.
Option 4
$\begin{aligned} P(A) + P(B) & = \text{0.7} \\ P(A \text{ and } B) & = \text{0.12} \\ \therefore P(A \text{ and } B) & \neq P(A) + P(B) \end{aligned}$

Therefore the events are not mutually exclusive.

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

If $P(A \text{ and } B) = 0$ then two events are mutually exclusive.

STEP: How do we decide if events are mutually exclusive?
[−1 point ⇒ 3 / 4 points left]

Mutually exclusive events can never occur at the same time. For example, "Akinbode is in the Western Cape" and "Akinbode is in Gauteng" are mutually exclusive events, because Akinbode cannot be in two places at once.

NOTE: Two events are mutually exclusive if the probability of both occuring at the same time is zero:

P (A and B) = 0

$P(A \text{ and } B) = 0$ .

STEP: Decide if the events are mutually exclusive
[−3 points ⇒ 0 / 4 points left]

For all events $A$ and $B$ :

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c} P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \end{array}$

There are two ways to prove that events are mutually exclusive:

Prove that $P(A \text{ and } B) = 0$ .
Prove that $P(A \text{ or } B) = P(A) + P(B)$ .

We know that:

${\color{red}{P(A) = \text{0.4}}}$
${\color{blue}{P(B) = \text{0.3}}}$ , and
${\color{ForestGreen}{P(A \text{ or } B) = \text{0.7}}}$ .

Method 1: Events $A$ and $B$ are mutually exclusive if $P(A \text{ and } B) = 0$

\begin{matrix} 0.7 = 0.4 + 0.3 - P (A and B) \end{matrix}

$\begin{array}{c} {\color{ForestGreen}{\text{0.7}}} = {\color{red}{\text{0.4}}} + {\color{blue}{\text{0.3}}} - P(A \text { and } B) \end{array}$

\begin{aligned} ∴ P (A and B) & = 0 \end{aligned}

$\begin{aligned} \therefore P(A \text { and } B) & = \text{0} \end{aligned}$

Therefore the events are mutually exclusive.

Method 2: Events $A$ and $B$ are mutually exclusive if $P(A \text{ or } B) = P(A) + P(B)$

\begin{aligned} P (A) + P (B) & = 0.4 + 0.3 \\ = 0.7 \\ ∴ P (A) + P (B) & = P (A or B) \end{aligned}

$\begin{aligned} P(A) + P(B) & = {\color{red}{\text{0.4}}} + {\color{blue}{\text{0.3}}} \\ & = \text{0.7} \\ \therefore P(A) + P(B) & = P(A \text{ or } B) \end{aligned}$

Therefore the events are mutually exclusive.

NOTE: Choose the method that you prefer, and use that one. You don't need to use both methods!

Of the given choices, Option 2 is correct.

Submit your answer as:

Option 2

\begin{aligned} P (A) + P (B) & = 0.7 \\ ∴ P (A) + P (B) & = P (A or B) \end{aligned}

$\begin{aligned} P(A) + P(B) & = \text{0.7} \\ \therefore P(A) + P(B) & = P(A \text{ or } B) \end{aligned}$

Therefore the events are mutually exclusive.

ID is: 4344 Seed is: 8513

Prove that two events are mutually exclusive

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

Given two events $A$ and $B$ :

$P(A) = \text{0.3}$
$P(B) = \text{0.5}$
$P(A \text{ or } B) = \text{0.8}$

Are the events $A$ and $B$ mutually exclusive? Justify your answer with appropriate calculations.

INSTRUCTION: Choose the most correct option to answer this question.

Answer:

Option 1
$\begin{aligned} P(A) + P(B) & = \text{0.8} \\ \therefore P(A) + P(B) & = P(A \text{ or } B) \end{aligned}$

Therefore the events are mutually exclusive.
Option 2
$\begin{aligned} P(A \text{ and } B) & = \text{0.15} \\ \therefore P(A \text{ and } B) & = P(A) \times P(B) \end{aligned}$

Therefore the events are mutually exclusive.
Option 3
$\begin{aligned} P(A) \times P(B) & = \text{0.15} \\ \therefore P(A) \times P(B) & \neq 0 \end{aligned}$

Therefore the events are not mutually exclusive.
Option 4
$\begin{aligned} P(A) + P(B) & = \text{0.8} \\ P(A \text{ and } B) & = \text{0.15} \\ \therefore P(A \text{ and } B) & \neq P(A) + P(B) \end{aligned}$

Therefore the events are not mutually exclusive.

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

If $P(A \text{ and } B) = 0$ then two events are mutually exclusive.

STEP: How do we decide if events are mutually exclusive?
[−1 point ⇒ 3 / 4 points left]

Mutually exclusive events can never occur at the same time. For example, "Bukelwa is in the Western Cape" and "Bukelwa is in Gauteng" are mutually exclusive events, because Bukelwa cannot be in two places at once.

NOTE: Two events are mutually exclusive if the probability of both occuring at the same time is zero:

P (A and B) = 0

$P(A \text{ and } B) = 0$ .

STEP: Decide if the events are mutually exclusive
[−3 points ⇒ 0 / 4 points left]

For all events $A$ and $B$ :

\begin{matrix} P (A or B) = P (A) + P (B) - P (A and B) \end{matrix}

$\begin{array}{c} P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \end{array}$

There are two ways to prove that events are mutually exclusive:

Prove that $P(A \text{ and } B) = 0$ .
Prove that $P(A \text{ or } B) = P(A) + P(B)$ .

We know that:

${\color{red}{P(A) = \text{0.3}}}$
${\color{blue}{P(B) = \text{0.5}}}$ , and
${\color{ForestGreen}{P(A \text{ or } B) = \text{0.8}}}$ .

Method 1: Events $A$ and $B$ are mutually exclusive if $P(A \text{ and } B) = 0$

\begin{matrix} 0.8 = 0.3 + 0.5 - P (A and B) \end{matrix}

$\begin{array}{c} {\color{ForestGreen}{\text{0.8}}} = {\color{red}{\text{0.3}}} + {\color{blue}{\text{0.5}}} - P(A \text { and } B) \end{array}$

\begin{aligned} ∴ P (A and B) & = 0 \end{aligned}

$\begin{aligned} \therefore P(A \text { and } B) & = \text{0} \end{aligned}$

Therefore the events are mutually exclusive.

Method 2: Events $A$ and $B$ are mutually exclusive if $P(A \text{ or } B) = P(A) + P(B)$

\begin{aligned} P (A) + P (B) & = 0.3 + 0.5 \\ = 0.8 \\ ∴ P (A) + P (B) & = P (A or B) \end{aligned}

$\begin{aligned} P(A) + P(B) & = {\color{red}{\text{0.3}}} + {\color{blue}{\text{0.5}}} \\ & = \text{0.8} \\ \therefore P(A) + P(B) & = P(A \text{ or } B) \end{aligned}$

Therefore the events are mutually exclusive.

NOTE: Choose the method that you prefer, and use that one. You don't need to use both methods!

Of the given choices, Option 1 is correct.

Submit your answer as:

Option 1

\begin{aligned} P (A) + P (B) & = 0.8 \\ ∴ P (A) + P (B) & = P (A or B) \end{aligned}

$\begin{aligned} P(A) + P(B) & = \text{0.8} \\ \therefore P(A) + P(B) & = P(A \text{ or } B) \end{aligned}$

Therefore the events are mutually exclusive.

ID is: 3140 Seed is: 150

Mutually exclusive events: phone numbers

In Pretoria phone numbers start with $\text{012}$ . After that there are seven more digits:

$\text{012}$ -XXX-XXXX

With these seven digits, there are many phone numbers possible in Pretoria. Here are two statements about one of these seven-digit phone numbers:

The first digit in the phone number is $\text{3}$ .
The fifth and the seventh digits are both $\text{7}$ .

Answer: Are the statements mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two things are mutually exclusive if they cannot both be true at the same time. Can both statements be true about a single phone number?

STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

The words 'mutually exclusive' mean that two things cannot both be true at the same time: if one of them is true the other must be false. In other words, there is a conflict between the statements. For example: 'The ANC has the majority in Parliament' and 'The IFP has the majority in Parliament.' If one of these statements is true, the other must be false. They are mutually exclusive statements because there is a conflict.

To decide if the statements are mutually exclusive, we must imagine one phone number which is seven digits long and ask ourselves: if one of the statements is true about that phone number, does that make the other statement false?

These statements are not mutually exclusive. We can even write down an example which agrees with both statements: 338-6737. If it is possible to write one phone number which agrees with both statements, they cannot be mutually exclusive.

The correct choice from the list is: No.

Submit your answer as:

ID is: 3140 Seed is: 4390

Mutually exclusive events: phone numbers

In Nelspruit phone numbers start with $\text{013}$ . After that there are seven more digits:

$\text{013}$ -XXX-XXXX

With these seven digits, there are many phone numbers possible in Nelspruit. Here are two statements about one of these seven-digit phone numbers:

The phone number has the digit $\text{7}$ six times.
The seventh digit in the phone number is $\text{4}$ .

Answer: Are the statements mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two things are mutually exclusive if they cannot both be true at the same time. Can both statements be true about a single phone number?

STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

The words 'mutually exclusive' mean that two things cannot both be true at the same time: if one of them is true the other must be false. In other words, there is a conflict between the statements. For example: 'The ANC has the majority in Parliament' and 'The DA has the majority in Parliament.' If one of these statements is true, the other must be false. They are mutually exclusive statements because there is a conflict.

These statements are not mutually exclusive. We can even write down an example which agrees with both statements: 777-7774. If it is possible to write one phone number which agrees with both statements, they cannot be mutually exclusive.

The correct choice from the list is: No.

Submit your answer as:

ID is: 3140 Seed is: 1699

Mutually exclusive events: phone numbers

In Pretoria phone numbers start with $\text{012}$ . After that there are seven more digits:

$\text{012}$ -XXX-XXXX

With these seven digits, there are many phone numbers possible in Pretoria. Here are two statements about one of these seven-digit phone numbers:

The phone number includes the digit 8 only once.
The phone number does not include the digit 8.

Answer: Are the statements mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two things are mutually exclusive if they cannot both be true at the same time. Can both statements be true about a single phone number?

STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

These statements are mutually exclusive: they conflict with each other. No phone number can include a $\text{8}$ and at the same time not include a $\text{8}$ . (If one of the statements is true, the other must be false.)

The correct choice from the list is: Yes.

Submit your answer as:

ID is: 3136 Seed is: 3936

Mutually exclusive events

Answer the two questions below about mutually exclusive events.

What does it mean if two events are 'mutually exclusive?'
Answer: Mutually exclusive means: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive statements conflict with each other. Here is an example of mutually exclusive statements; use them to choose the correct answer.
1. My name starts with the letter A.
2. My name starts with the letter K.
STEP: Identify the statement which means the same thing as "mutually exclusive"
[−1 point ⇒ 0 / 1 points left]

The words "mutually exclusive" refer to two things that exclude each other. They cannot both be true at the same time. For example: "The ANC has the majority in Parliament" and "The IFP has the majority in Parliament." If one of these statements is true, the other must be false. They are mutually exclusive statements.

The choice from the list which has the same meaning as this is: there is no chance that the events can happen together.

Submit your answer as:
Think about a person going into a shop where he can buy airtime and other basic things like milk, juice and chips. The following statements might describe what happens in the shop:
- The person buys a single airtime voucher.
- The person buys airtime from Virgin Mobile and Cell C.
Answer: Are the statements mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Ask yourself this: can both of the statements be true when the person is in the shop? If the answer is no, then the statements are mutually exclusive.

STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

As stated above, two things are mutually exclusive if they cannot both be true at the same time. To decide if the statements are mutually exclusive, we must imagine that the person is in the shop and ask ourselves: if one of the statements is true, does that mean the other statement must be false? If so, the statements are mutually exclusive.

These statements are mutually exclusive because they exclude each other. If the person only bought one airtime voucher, it cannot also be true that they bought airtime from $\text{2}$ different service providers.

The correct choice from the list is: Yes.

Submit your answer as:

ID is: 3136 Seed is: 274

Mutually exclusive events

Answer the two questions below about mutually exclusive events.

What does it mean if two events are 'mutually exclusive?'
Answer: Mutually exclusive means: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive statements conflict with each other. Here is an example of mutually exclusive statements; use them to choose the correct answer.
1. My name starts with the letter A.
2. My name starts with the letter K.
STEP: Identify the statement which means the same thing as "mutually exclusive"
[−1 point ⇒ 0 / 1 points left]

The words "mutually exclusive" refer to two things that exclude each other. They cannot both be true at the same time. For example: "The ANC has the majority in Parliament" and "The EFF has the majority in Parliament." If one of these statements is true, the other must be false. They are mutually exclusive statements.

The choice from the list which has the same meaning as this is: that two things cannot occur at the same time.

Submit your answer as:
Think about a person going into a shop where she can buy airtime and other basic things like milk, juice and chips. The following statements might describe what happens in the shop:
- The person buys a single airtime voucher.
- The person buys airtime from Vodacom and Telkom Mobile.
Answer: Are the statements mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Ask yourself this: can both of the statements be true when the person is in the shop? If the answer is no, then the statements are mutually exclusive.

STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

As stated above, two things are mutually exclusive if they cannot both be true at the same time. To decide if the statements are mutually exclusive, we must imagine that the person is in the shop and ask ourselves: if one of the statements is true, does that mean the other statement must be false? If so, the statements are mutually exclusive.

These statements are mutually exclusive because they exclude each other. If the person only bought one airtime voucher, it cannot also be true that they bought airtime from $\text{2}$ different service providers.

The correct choice from the list is: Yes.

Submit your answer as:

ID is: 3136 Seed is: 5813

Mutually exclusive events

Answer the two questions below about mutually exclusive events.

What does it mean if two events are 'mutually exclusive?'
Answer: Mutually exclusive means: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive statements conflict with each other. Here is an example of mutually exclusive statements; use them to choose the correct answer.
1. My name starts with the letter A.
2. My name starts with the letter K.
STEP: Identify the statement which means the same thing as "mutually exclusive"
[−1 point ⇒ 0 / 1 points left]

The words "mutually exclusive" refer to two things that exclude each other. They cannot both be true at the same time. For example: "The ANC has the majority in Parliament" and "The EFF has the majority in Parliament." If one of these statements is true, the other must be false. They are mutually exclusive statements.

The choice from the list which has the same meaning as this is: that two things cannot occur at the same time.

Submit your answer as:
Think about a bus moving along toward a bus stop. The following statements describe what might happen when the bus comes to the bus stop:
- The bus does not stop.
- $\text{4}$ people get off the bus.
Answer: Are the statements mutually exclusive?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Ask yourself this: can both of the statements be true when the bus comes to a stop? If the answer is no, then the statements are mutually exclusive.

STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

As stated above, two things are mutually exclusive if they cannot both be true at the same time. To decide if the statements are mutually exclusive, we must imagine that the bus comes to a stop and ask ourselves: if one of the statements is true, does that mean the other statement must be false? If so, the statements are mutually exclusive.

These statements are mutually exclusive because they exclude each other. If the bus does not stop, it cannot be true that 4 people got off the bus.

The correct choice from the list is: Yes.

Submit your answer as:

MONTHLY INCOME (IN RANDS)	Area 1	Area 2	TOTAL
$x < \text{3,200}$	$\text{1,867}$	$\text{800}$	$\text{2,667}$
$\text{3,200} \leq x < \text{25,600}$	$\text{1,444}$	$\text{438}$	$\text{1,882}$
$x \geq \text{25,600}$	$\text{84}$	$\text{218}$	$\text{302}$
Total	$\text{3,395}$	$\text{1,456}$	$\text{4,851}$

MONTHLY INCOME (IN RANDS)	Area 1	Area 2	TOTAL
$x < \text{3,200}$	$\text{1,867}$	$\text{800}$	$\text{2,667}$
$\text{3,200} \leq x < \text{25,600}$	$\text{1,444}$	$\text{438}$	$\text{1,882}$
$x \geq \text{25,600}$	$\text{84}$	$\text{218}$	$\text{302}$
Total	$\text{3,395}$	$\text{1,456}$	$\text{4,851}$

MONTHLY INCOME (IN RANDS)	Area 1	Area 2	TOTAL
$x < \text{3,200}$	$\text{1,433}$	$\text{1,347}$	$\text{2,780}$
$\text{3,200} \leq x < \text{25,600}$	$\text{717}$	$\text{1,079}$	$\text{1,796}$
$x \geq \text{25,600}$	$\text{55}$	$\text{269}$	$\text{324}$
Total	$\text{2,205}$	$\text{2,695}$	$\text{4,900}$

MONTHLY INCOME (IN RANDS)	Area 1	Area 2	TOTAL
$x < \text{3,200}$	$\text{1,433}$	$\text{1,347}$	$\text{2,780}$
$\text{3,200} \leq x < \text{25,600}$	$\text{717}$	$\text{1,079}$	$\text{1,796}$
$x \geq \text{25,600}$	$\text{55}$	$\text{269}$	$\text{324}$
Total	$\text{2,205}$	$\text{2,695}$	$\text{4,900}$

MONTHLY INCOME (IN RANDS)	Area 1	Area 2	TOTAL
$x < \text{3,200}$	$\text{427}$	$\text{190}$	$\text{617}$
$\text{3,200} \leq x < \text{25,600}$	$\text{856}$	$\text{666}$	$\text{1,522}$
$x \geq \text{25,600}$	$\text{1,568}$	$\text{1,046}$	$\text{2,614}$
Total	$\text{2,851}$	$\text{1,902}$	$\text{4,753}$

A	$P(C \text{ or } D)$	$P(C) \times P(D)$
B	$P(C \text{ or } D)$	$P(C) + P(D)$
C	$P(C \text{ and } D)$	$P(C) + P(D)$
D	$P(C \text{ and } D)$	$P(C) \times P(D)$

A	$P(M \text{ and } N)$	$P(M) \times P(N)$
B	$P(M \text{ or } N)$	$P(M) \times P(N)$
C	$P(M \text{ and } N)$	$P(M) + P(N)$
D	$P(M \text{ or } N)$	$P(M) + P(N)$

A	$P(G \text{ and } H)$	$P(G) + P(H)$
B	$P(G \text{ or } H)$	$P(G) \times P(H)$
C	$P(G \text{ or } H)$	$P(G) + P(H)$
D	$P(G \text{ and } H)$	$P(G) \times P(H)$

	Chicken	Beef	Totals
Rice	$\text{95}$	$\text{31}$	$\text{126}$
Pasta	$a$	$\text{46}$	$\text{65}$
Totals	$\text{114}$	$b$	$c$

	Accounting	History	Totals
11	$a$	$b$	$c$
12	$\text{83}$	$\text{42}$	$\text{125}$
Totals	$\text{91}$	$\text{91}$	$\text{182}$

	Females	Males	Total
Positive	A:	B:	C:
Not Positive	D:	E:	F:
Total	G:	H:	Total:

	Year 9	Year 6	Total
a packed lunch	A:	B:	C:
Not a packed lunch	D:	E:	F:
Total	G:	H:	Total:

	Year 8	Year 11	Total
Science	A:	B:	C:
Not Science	D:	E:	F:
Total	G:	H:	Total:

	WATCHED A MOVIE DURING EXAMS	DID NOT WATCH A MOVIE DURING EXAMS	TOTALS
Male	$\text{27}$	$\text{33}$	$\text{60}$
Female	$\text{48}$	$\text{72}$	$\text{120}$
Totals	$\text{75}$	$\text{105}$	$\text{180}$

	PLAYED VIDEO GAMES DURING EXAMS	DID NOT PLAY VIDEO GAMES DURING EXAMS	TOTALS
Male	$\text{61}$	$\text{19}$	$\text{80}$
Female	$\text{67}$	$\text{33}$	$\text{100}$
Totals	$\text{128}$	$\text{52}$	$\text{180}$

	Dog	Cat	Total
Girls	x	y	z
Boys	56	11	67
Total	104	100	204

	Coke	Fanta	Total
Girls	60	5	65
Boys	8	z	59
Total	x	y	124

	Dog	Cat	Total
Girls	57	24	z
Boys	33	59	y
Total	x	83	173

MONTHLY INCOME (IN RANDS)	Area 1	Area 2	TOTAL
$\color{black}{x < \text{3,200}}$	$\color{black}{\text{366}}$	$\color{gray}{\text{40}}$	$\color{gray}{\text{406}}$
$\text{3,200} \leq x < \text{25,600}$	$\color{black}{\text{732}}$	$\color{gray}{\text{1,018}}$	$\color{gray}{\text{1,750}}$
$\color{ForestGreen}{x \geq \text{25,600}}$	$\color{black}{\text{1,342}}$	$\color{gray}{\text{569}}$	$\color{gray}{\text{1,911}}$
Total	$\color{black}{\text{2,440}}$	$\color{gray}{\text{1,627}}$	$\color{gray}{\text{4,067}}$

Probability

1. Revision of probability

Probability basics

Probability basics

Probability basics

Using the probability identity

Using the probability identity

Using the probability identity

Exploring definitions

Exploring definitions

Exploring definitions

Relative frequency compared to theoretical probability

Relative frequency compared to theoretical probability

Relative frequency compared to theoretical probability

Theoretical probability

Theoretical probability

Theoretical probability

Applying theoretical probability to experimental trials

Applying theoretical probability to experimental trials

Applying theoretical probability to experimental trials

Revision: calculations with probability

Revision: calculations with probability

Revision: calculations with probability

Relative frequency

Relative frequency

Relative frequency

Probability identities

Probability identities

Probability identities

2. Equiprobable sample space

3. Mutually exclusive events

Mutually exclusive events

Mutually exclusive events

Mutually exclusive events

Mutually exclusive sets

Mutually exclusive sets

Mutually exclusive sets

Prove that two events are mutually exclusive

Prove that two events are mutually exclusive

Prove that two events are mutually exclusive

Mutually exclusive events: phone numbers

Mutually exclusive events: phone numbers

Mutually exclusive events: phone numbers

Mutually exclusive events

Mutually exclusive events

Mutually exclusive events

4. Complementary events

Complementary sets: basic facts

Complementary sets: basic facts

Complementary sets: basic facts

Complementary sets: identifying a complementary set

Complementary sets: identifying a complementary set

Complementary sets: identifying a complementary set

Venn diagram: complementary sets

Venn diagram: complementary sets

Venn diagram: complementary sets

Working with complementary events

Working with complementary events

Working with complementary events

Complementary sets: working with set notation

Complementary sets: working with set notation

Complementary sets: working with set notation

5. Independent events

Contingency tables: proving whether events are independent

Contingency tables: proving whether events are independent

Contingency tables: proving whether events are independent

Dependent and independent events

Dependent and independent events

Dependent and independent events

6. Practical applications

Two-way tables

Two-way tables

Two-way tables

Tree diagrams with replacement

Tree diagrams with replacement

Tree diagrams with replacement

Contingency tables

Contingency tables

Contingency tables

Tree diagrams with replacement